The complete balanced chemical reactions are:
HNO3 => CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2(g)
H2SO4 => CaCO3 + H2SO4 → CaSO4 + H2O + CO2(g)
So we see that 1 mole of CaCO3 is needed for 2 moles of HNO3 and similarly
to 1 mole of H2SO4.
The number of moles can be calculated as the product of
volume and molarity, so:
moles H2SO4 = 1.7×10^−5 M * (15.5 x 10^9 L) = 263,500 mol H2SO4
moles HNO3 = 8.9×10^−6 M * (15.5 x 10^9 L) = 137,950 mol HNO3
So the total moles of CaCO3 required is:
moles CaCO3 = 263,500 mol * 1 + 137,950 mol * (1/2)
moles CaCO3 = 332,475 mol
The molar mass of CaCO3 is 100.086 g/mol, so the mass is:
mass CaCO3 = 332,475 mol * 100.086 g/mol
mass CaCO3 = 33,276,092.85 g = 33.3 x 10^3 kg
Mass of limestone required for complete neutralization is 3,336.64 [tex]\rm \times\;10^4[/tex] grams.
Limestone is [tex]\rm CaCO_3[/tex]. The neutralization reaction for limestone will be:
[tex]\rm CaCO_3\;+\;2\;HNO_3\;\rightarrow\;Ca(NO_3)_2\;+\;H_2O\;+\;CO_2[/tex]
[tex]\rm CaCO_3\;+\;H_2SO_4\;\rightarrow\;CaSO_4\;+\;H_2O\;+\;CO_2[/tex]
The balanced equation states that for neutralization of 1 mole of limestone, 2 moles [tex]\rm HNO_3[/tex] and 1 mole of [tex]\rm H_2SO_4[/tex] is required.
The moles of [tex]\rm HNO_3[/tex] and [tex]\rm H_2SO_4[/tex] available are:
Moles of [tex]\rm HNO_3[/tex] = molarity [tex]\times[/tex] volume (L)
Moles of [tex]\rm HNO_3[/tex] = 8.9 [tex]\rm \times\;10^-^6[/tex] [tex]\times[/tex] 15.5 [tex]\rm \times\;10^9[/tex] L
Moles of [tex]\rm HNO_3[/tex] = 13.975 [tex]\rm \times\;10^4[/tex] moles
Moles of [tex]\rm H_2SO_4[/tex] = 1.7 [tex]\rm \times\;10^-^5[/tex] [tex]\times[/tex] 15.5 [tex]\rm \times\;10^9[/tex] L
Moles of [tex]\rm H_2SO_4[/tex] = 26.35 [tex]\rm \times\;10^4[/tex] moles
Moles of [tex]\rm CaCO_3[/tex] required = 1 mole of [tex]\rm H_2SO_4[/tex] + [tex]\rm \dfrac{1}{2}[/tex] moles of [tex]\rm HNO_3[/tex]
Moles of [tex]\rm CaCO_3[/tex] required = 26.35 [tex]\rm \times\;10^4[/tex] moles [tex]\rm H_2SO_4[/tex] + [tex]\rm \dfrac{1}{2}[/tex] (13.975 [tex]\rm \times\;10^4[/tex] moles) [tex]\rm HNO_3[/tex]
Moles of [tex]\rm CaCO_3[/tex] required = 26.35 [tex]\rm \times\;10^4[/tex] + 6.9875 [tex]\rm \times\;10^4[/tex]
Moles of [tex]\rm CaCO_3[/tex] required = 33.3375 [tex]\rm \times\;10^4[/tex] moles
Mass of limestone = moles of limestone [tex]\times[/tex] molecular weight of limestone
Mass of limestone = 33.3375 [tex]\rm \times\;10^4[/tex] moles [tex]\times[/tex] 100.0869 grams
Mass of limestone = 3,336.64 [tex]\rm \times\;10^4[/tex] grams.
For more information refer the link;
https://brainly.com/question/3733001?referrer=searchResults
So the total moles of CaCO3 required is:
moles CaCO3 = 263,500 mol * 1 + 137,950 mol * (1/2)
moles CaCO3 = 332,475 mol
The molar mass of CaCO3 is 100.086 g/mol, so the mass is:
mass CaCO3 = 332,475 mol * 100.086 g/mol
mass CaCO3 = 33,276,092.85 g = 33.3 x 10^3 kg
